Cantor Confusion
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From: Virgil on 30 Oct 2006 16:54 In article <1162216575.579280.178630(a)m73g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > David Marcus schrieb: > > > mueckenh(a)rz.fh-augsburg.de wrote: > > > MoeBlee schrieb: > > > > David Marcus wrote: > > > > > Not sure if he ever said precisely "within Z set theory", but he > > > > > certainly said things very similar. Below are a few messages that I > > > > > found. There are probably others. > > > > > > > > > > In the first, he says that "standard mathematics contains a > > > > > contradiction". > > > > > > as ar as standard mathematics is derived from set theory an comes to > > > the conclusion of an empty vase at noon or an uncountable set of reals. > > > Classical mathematics is free of such contradictions. > > > > > > > >In the next two, he states there are "internal > > > > > contradictions of set theory". In the next, I say that he says that > > > > > "standard mathematics contains a contradiction", and he does not > > > > > dispute this. > > > > > > > > Thanks. And at least a couple of times I said that I was reading his > > > > argument to see whether it does sustain his claim about set theory, as > > > > I mentioned specifically Z set theory. > > > > > > My proof of the binary tree covers all possible theories. And it should > > > not cost you too much time to see that it is true. > > > > Unfortunately, I don't know what "covers all possible theories" means. > > (You seem to use the word "cover" a lot.) > > Here it means, the result of my proof is valid for all possible > theories which allow to define the infinite binary tree. Except for the actual theory of infinite binary trees, which WM does not comprehend. > > > Either you have a proof that can be given in ZFC or you don't. Which is > > it? This shouldn't be a difficult question to answer. > > But it is not an interesting question. We find it very interesting that WM will not, or more likely cannot, answer it.
From: David Marcus on 30 Oct 2006 16:55 mueckenh(a)rz.fh-augsburg.de wrote: > Sebastian Holzmann schrieb: > > mueckenh(a)rz.fh-augsburg.de <mueckenh(a)rz.fh-augsburg.de> wrote: > > > Every member of a list of only finite sequences is a finite sequence. > > > Every diagonal of such a list is a finite sequence. That is enough. > > > > That is wrong. > > All entries of the list have a finite number of letters. An infinite > sequence is larger than any finite sequence. The diagonal of a list > cannot have more letters than the lines. > > According to your logic the list can have infinitely many lines. But > even if that was correct it would not facilitate an infinte diagonal. > > The number of diagonal elements is the minimum of columns and lines. OK, let's use our logic (or standard logic or ZFC). The list has infinitely many lines and columns, so the number of diagonal elements is infinite. Your statement that the number of diagonal elements is finite is wrong. You've jumped from "each entry is finite" to "the number of columns is finite". -- David Marcus
From: David Marcus on 30 Oct 2006 16:59 mueckenh(a)rz.fh-augsburg.de wrote: > William Hughes schrieb: > > > Let our alphebet be {0,1,2}. Let our diagonal construction be > > 0->1, 1->2, 2->0. Define a finite sequence as one that has only 0's > > after a certain point. The set A only has sequences that have only 0's > > after a certain point. > > > > A begins > > > > 000... > > 1000... > > 2000... > > 11000... > > 12000... > > > > The diagonal is an unending string of ones. The set A does not contain > > the diagonal. > > > > > > Thus there is no contradiction. You need a diagaonal before > > > > you can get a contradiction, therefore you need set B. > > > > > > Why should we not construct the diagonal of these sequeces (words) of > > > A? > > > > We can do this but the diagonal is not a finite sequence, so it is > > not a member of A. > > If the list consists of finite sequences, then the diaogonal is a > finite sequence too. Because it cannot be broader than the list In ZFC or in some other system? > All entries of the list have a finite number of letters. An infinite > sequence is larger than any finite sequence. The diagonal of a list > cannot have more letters than the lines. -- David Marcus
From: Virgil on 30 Oct 2006 17:00 In article <1162217114.958400.100140(a)e64g2000cwd.googlegroups.com>, mueckenh(a)rz.fh-augsburg.de wrote: > Virgil schrieb: > > > > > Forget the Turing machines. The diagonal of the list of all finite > > > sequences (words) of a finite alphabet is a finite sequence because the > > > diagonal cannot have more places than the words in the list. > > > > > > But if the nth word has, say, n+1 places, then one has a list of finite > > sequences which are unbounded. > > > > There is a vital difference between a list of merely finite sequences > > and a list of uniformly bounded sequences ( all being bounded by the > > same bound) which allow the former to have sequences of arbitrarily > > large, but still finite, lengths. > > Every member of a list of only finite sequences is a finite sequence. But that does not limit the number of such sequences to being finite. > Every diagonal of such a list is a finite sequence. Wrong! Define the infinite sequence of finite strings in which the nth string consists of n 1's, and define the infinite diagonal so that in its nth position it has a "2" for each of infinitely many n's. Then this example demonstrates WM's wrongness.
From: Lester Zick on 30 Oct 2006 17:06
On 30 Oct 2006 12:28:41 -0800, "MoeBlee" <jazzmobe(a)hotmail.com> wrote: >Lester Zick wrote: >> You mean the question where you demand I research and justify your >> opinions for you, Moe? > >I didn't demand anything of you in a question, let alone that I've >never demanded of you that you justify my opinions. My mistake, Moe. It's just that you keep snipping all these posts. Makes it kinda hard to tell who demanded what. Must have been I who demanded I justify your opinions. Maybe I've been snipping all these posts too. ~v~~ |